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1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(\frac{121212}{161616}-\left(\frac{151515}{323232}-x\right)=2\)
=> \(\frac{3}{4}-\left(\frac{15}{32}-x\right)=2\)
=> \(\frac{15}{32}-x=\frac{3}{4}-2\)
=> \(\frac{15}{32}-x=-\frac{5}{4}\)
=> \(x=\frac{15}{32}-\frac{-5}{4}=\frac{15}{32}+\frac{5}{4}=\frac{55}{32}\)
b) \(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}+\frac{x}{90}=\frac{9}{5}\)
=> \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}+\frac{x}{6\cdot7}+\frac{x}{7\cdot8}+\frac{x}{8\cdot9}+\frac{x}{9\cdot10}=\frac{9}{5}\)
=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{9}-\frac{x}{10}=\frac{9}{5}\)
=> \(\frac{x}{1}-\frac{x}{10}=\frac{9}{5}\)
=> \(\frac{10x-x}{10}=\frac{9}{5}\)
=> \(\frac{9x}{10}=\frac{9}{5}\)
=> \(\frac{9x}{10}=\frac{9\cdot2}{5\cdot2}=\frac{18}{10}\)
=> x = 2
\(\frac{121}{35}.\left(\frac{1}{7}-\frac{5}{21}\right).\left|-\frac{11}{7}\right|\)
\(=\frac{121}{35}.\left(\frac{3}{21}-\frac{5}{21}\right).\frac{11}{7}\)
\(=\frac{121}{35}.\frac{-2}{21}.\frac{11}{7}\)
\(=-\frac{242}{735}.\frac{11}{7}\)
\(=-\frac{2662}{5145}\)
Chúc bạn học tốt !!!
Xem lại đề nhé !!! Số to quá !!!
\(\frac{121}{35}\times\left(\frac{1}{7}-\frac{5}{21}\right)\times\left|\frac{-11}{7}\right|\)
\(=\frac{121}{35}\times\frac{-2}{21}\times\frac{11}{7}\)
\(=\frac{-2662}{5145}\)
Chúc bn học tốt !!!
Mk định ko tl nhưng thấy ''đề sai'' mk nghĩ bn sai chứ để x là PS ko sai
\(\frac{-2}{3}+\frac{x}{5}=\frac{1}{5}\)
\(\frac{-10}{15}+\frac{3x}{15}=\frac{3}{15}\)
\(-10+3x=3\)
\(3x=3-\left(-10\right)\Leftrightarrow3x=13\Leftrightarrow x=\frac{13}{3}\)
Bài 6:
a/ \(x\in\left\{-2;-1\right\}\)
b/ \(x\in\left\{-4;-3;-2\right\}\)
c/ \(x\in\left\{-5;-4\right\}\)
d/ \(x\in\left\{-1;0;1;2;3\right\}\)
e/ \(x\in\left\{7,8,9\right\}\)
g/ \(x\in\left\{-6;-5;-4;-3;-2;-1\right\}\)
Bài 7:
a/ \(-26+\left(-32\right)=-\left(26+32\right)=-58\)
b/ \(-267+\left(-473\right)=-\left(267+473\right)=-740\)
c/ \(27+\left(-43\right)=-\left(43-27\right)=-16\)
d/ \(126+\left(-34\right)=126-34=92\)
e/ \(81+\left(-25\right)=81-25=56\)
g/ \(-92+\left(-62\right)=-\left(92+62\right)=-154\)
h/ \(-125+\left(-175\right)=-\left(125+175\right)=-300\)
i/ \(-34+\left(-26\right)=-\left(34+26\right)=-60\)
k/ \(-156+84=-\left(156-84\right)=-72\)
Ta có: 5+5\(^2\)+5\(^3\)+5\(^4\)+....+5\(^{60}\)= (5+5\(^2\))+(5\(^3\)+5\(^4\) ) +....+( 5\(^{59}\)+5\(^{60}\))=
= 30+ 5^2.(5+5^2)+...+5^58.(5+5^2)= 30+5^2.30+...+5^58.30= 30.(1+5^2+...+5^58)
Vì 30 \(⋮\)6 \(\Rightarrow\)30.(1+5^2+...+5^58) \(⋮\)6 hay 5+5\(^2\)+5\(^3\)+5\(^4\)+....+5\(^{60}\)\(⋮\)6
5+5\(^2\)+5\(^3\)+5\(^4\)+....+5\(^{60}\)= (5+5\(^2\)+5\(^3\) ) +(5\(^4\) + 5^5+5^6) +....+( 5^58+5\(^{59}\)+5\(^{60}\))=
= 155+ 5^3.(5+5^2+5^3)+...+5^57.(5+5^2+5^3)= 155+5^3.155+...+5^57.155=155.(1+5^3+...+5^57)
Vì 155 \(⋮\) 31 \(\Rightarrow\) 155.(1+5^3+...+5^57) \(⋮\) 31 hay 5+5\(^2\)+5\(^3\)+5\(^4\)+....+5\(^{60}\)\(⋮\) 31
Bạn vào chỗ câu hỏi của bạn Trương NGuyễn Ngọc Mỹ, giải tương tự giống bài của mình nhé
\(=143-43\cdot0=143\)
143-43.[(25:5)^2-5^2]
=143-43.(25-25)
=143-43.0
=143-0=143